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How many numbers n between 1 and 600 do exist such that the floor of $\sqrt[4]{n}$ divides n. please solved the problem in a general case.

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up vote 1 down vote accepted

Note that $\lfloor \sqrt[4]n \rfloor=2$ for $16 \le n \le 80$ so how many $n$'s in this range qualify?

If we let $k=\lfloor \sqrt[4] n \rfloor$ we have $\sum_{i=1}^{k-1}\left(\lfloor \frac {(i+1)^4-(i^4+2)}i\rfloor+1\right)+\lfloor \frac {n-(k^4+1)}k\rfloor+1$ for the general case. The idea is as the example above, to count the number of multiples of $i$ from $i^4$ through $(i+1)^4-1$. Note that we start and end with a multiple of $i$.

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There is 23 n that qualify in this range. – Emad Feb 4 '14 at 22:59
I updated the range-it should go through $80$, so $23$ is not correct. My fault. Since $\lfloor \sqrt[4]{600}\rfloor=4$ you only have three more ranges to worry about. – Ross Millikan Feb 4 '14 at 23:43

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